How to prove this series about Fibonacci number: $\sum_{n=1}^{\infty }\frac{F_{n}}{2^{n}}=2$? [duplicate]
How to prove this series result:
$$\sum_{n=1}^{\infty }\frac{F_{n}}{2^{n}}=2$$
where $F_{1}=1,~F_{2}=1,~F_n=F_{n-1}+F_{n-2},~~n\geq 3$.
I have no idea where to start.
Hint. You may use the fact that, if
$$f(x)=\sum_{n=0}^\infty F_nx^n \tag1$$
with $F_n$ the nth Fibonnacci number, $F_{n+2}=F_{n}+F_{n+1}$, then
$$f(x)={x\over 1-x-x^2}.\tag2$$
A proof of $(2)$ may be found here. Apply it to $x:=\dfrac12$.